Problem 81 - College Algebra with Modeling and Visualization

Question

A cardboard box with no top and a square base is being constructed and must have a volume of 108 cubic inches. Let \(x\) be the length of a side of its base in inches. (a) Write a formula \(A(x)\) that calculates the outside surface area in square feet of the box. (b) If cardboard costs \(\$ 0.10\) per square foot, write a formula \(C(x)\) that gives the cost in dollars of the cardboard in the box. (c) Find the dimensions of the box that would minimize the cost of the cardboard.

Step-by-Step Solution

Answer

The box dimensions are 6 inches by 6 inches for the base and 3 inches for the height.

Step 1: Define the Volume Constraint

Since the box is open on top and has a square base, the volume constraint can be expressed as \( V = x^2 \cdot h = 108 \), where \( h \) is the height of the box. We need to express \( h \) in terms of \( x \). Solving for \( h \), we have \( h = \frac{108}{x^2} \).

Step 2: Express the Surface Area Formula

The outside surface area of the box, \( A(x) \), includes the base area and the four sides. The base area is \( x^2 \), and the area of the four sides is \( 4xh \). So, the surface area \( A(x) = x^2 + 4x \cdot \frac{108}{x^2} \). Simplifying gives: \( A(x) = x^2 + \frac{432}{x} \).

Step 3: Convert Surface Area to Feet

The surface area formula \( A(x) \) is currently in square inches. There are 144 square inches in a square foot, so to convert \( A(x) \) to square feet, we divide by 144: \( A(x) = \frac{x^2}{144} + \frac{432}{144x} \).

Step 4: Calculate the Cost Formula

The cost formula \( C(x) \) is \( 0.10 \) multiplied by the surface area in square feet: \( C(x) = 0.10 \left( \frac{x^2}{144} + \frac{432}{144x} \right) \). Simplifying gives \( C(x) = \frac{x^2}{1440} + \frac{3}{10x} \).

Step 5: Find the Value of x that Minimizes the Cost

To minimize \( C(x) \), we need to find the derivative \( C'(x) \) and set it to zero. \( C(x) = \frac{x^2}{1440} + \frac{3}{10x} \) gives \( C'(x) = \frac{x}{720} - \frac{3}{10x^2} \). Setting \( C'(x) = 0 \) gives \( \frac{x}{720} = \frac{3}{10x^2} \). Solving for \( x \), \( x^3 = 216 \), so \( x = 6 \).

Step 6: Determine the Box Dimensions

With \( x = 6 \) inches, substitute back to find \( h \): \( h = \frac{108}{x^2} = \frac{108}{36} = 3 \). So, the dimensions are 6 inches by 6 inches for the base and 3 inches for the height.

Key Concepts

Volume Constraint

When designing a cardboard box with a specific volume, it's essential to understand the volume constraint concept. In this problem, the box is required to have a volume of 108 cubic inches, with a square base and no top.
Letting the side length of the square base be denoted as \( x \) and the height of the box as \( h \), the volume \( V \) can be described by the formula \( V = x^2 \, h \). Here, the challenge is to express the height in terms of the side length by rearranging the formula, giving us \( h = \frac{108}{x^2} \).
This equation helps us understand the relationship between the dimensions of the box and ensures that the volume remains constant at 108 cubic inches for any value of \( x \). Understanding such constraints is crucial for solving optimization problems.

Surface Area Calculation

The surface area of the cardboard box involves calculating the areas of all its visible surfaces, excluding the open top. Because the box has a square base, the base area is simply \( x^2 \).
However, to find the total surface area, we must also consider the four side walls. Each side wall has an area of \( x \, h \), where \( h = \frac{108}{x^2} \) is substituted from the earlier volume constraint formula, leading to a total side area of \( 4x \, h \).
By summing the base area and the total side area, we arrive at the formula for the outside surface area, \( A(x) = x^2 + \frac{432}{x} \). This formula allows us to understand how changes in \( x \) affect the overall surface area needed for the box.

Cost Minimization

To minimize the cost of constructing the cardboard box, we need to consider how the surface area affects costs directly, given that cardboard is priced at $0.10 per square foot.
First, convert the surface area from square inches to square feet by using the conversion factor: 144 square inches per square foot. The adjusted formula is \( A(x) = \frac{x^2}{144} + \frac{432}{144x} \).
Subsequently, the cost function, \( C(x) \), is obtained by multiplying the surface area function by the cost per square foot, yielding \( C(x) = 0.10 \left( \frac{x^2}{144} + \frac{432}{144x} \right) \). After simplification, the cost expression becomes \( C(x) = \frac{x^2}{1440} + \frac{3}{10x} \).
Optimizing this cost function requires finding the \( x \) that makes it smallest, which is a typical optimization problem.

Derivatives in Calculus

Derivatives play a key role in solving optimization problems such as cost minimization. To find the dimension size \( x \) that minimizes the cost, we need to compute the derivative of the cost function \( C(x) \).
The derivative, \( C'(x) \), is computed as \( \frac{x}{720} - \frac{3}{10x^2} \). Setting this derivative equal to zero, \( C'(x) = 0 \), provides the critical points where either a maximum or a minimum occurs.
After algebraically solving \( \frac{x}{720} = \frac{3}{10x^2} \), we find \( x = 6 \) inches as the value that minimizes the cardboard cost. This outcome illustrates how derivatives help in determining the optimal solution for such problems, highlighting their importance in calculus.